Can someone check my math here?

Discussion in 'General Electronics Chat' started by magnet18, Feb 21, 2012.

  1. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    Alright, I need to take a 5V power source, and power a 3V EL wire unit with it through a transistor.
    The EL wire unit draws 115mA, and the transistor is a BC337, 100 gain, the base is fed from a 5V pin off an arduino.

    5-3=2V drop, 2/.115=17.4 Ω for the collector resistor? (do I even need that, it's so small?)
    17.4*100 gain = 1.74KΩ for the base resistor?
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Like most transistors, Vce(sat) for BC337 is defined when Ib=Ic/10. So Ib≈11.5mA
    Rb=(Vin-Vbe).Ib=(5-0.7)/11.5mA,
    Rb=374Ω
    I would use 330Ω to account for the Arduino output resistance.
     
  3. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Use the BC337 with its collector connected to the 5v rail, and drive it (as emitter follower) with a voltage divider that puts approx 3.7v on its base (when the micro pin is HI).

    That way the BC337 will make a loosely regulated 3v at its emitter, which is now the "3v power supply" going to the EL module.

    the BC337 will have a gain easily well over 200 when supplying 115mA, so you only need 3 or 4 mA into its base.
     
  4. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    Thank you both, though I think I might just go with some MOSFETS and keep it simple.
    All I have laying around is 250V 20A... should work ;)
     
  5. THE_RB

    AAC Fanatic!

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    If the use the BC337 and two resistors in the configuration I mentioned it provides a benefit you can't get from a mosfet, ie it acts as a 3v regulator which is switched on/off by your micro.
     
  6. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    How does that work?
     
  7. Ron H

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    Apr 14, 2005
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    Here is a simulation.
     
  8. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    Thanks :)
    I'm still not sure why putting 3.7V on the base limits it to 3V at the emitter though?
    Is it because it is "locked" at 3.7V, and the cb portion of the transistor dissipates the 2V of energy, leaving 3V at the emitter after the .7V drop of the be portion of the transistor?

    (also, would it work if I put 6V on the collector?)
     
  9. Ron H

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    Apr 14, 2005
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    It's an emitter follower (common collector amplifier). The output follows the base, with the emitter being ≈0.7V lower than the base (0.7V higher for a PNP). The base current will be equal to the emitter current divided by β+1. So, we set up a voltage divider on the input, reducing 5V to ≈3.7V. The equivalent resistance of the divider (R2||R3) must be low enough that the base current (which is poorly defined because β varies with current, and from unit to unit) does not substantially affect the voltage on the base. When the input is zero, the emitter will also be zero, because it can't go any lower (unless you had a negative supply on the other end of the emitter load). The transistor will be off in this case.
    The collector voltage can be any value above ≈4V, up to the value that exceeds the Vce(max) specs on the transistor, or causes it to overheat due to power dissipation (Ic*Vce). Beta does vary a little as Vce varies, tending to be higher for higher values of Vce.
     
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  10. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    Alright, thanks, but I'm still wondering how the 3volts is dropped from the collector to the base
     
  11. Ron H

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    Apr 14, 2005
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    It's because if the emitter were to go to less than 0.7V below the base, the transistor would begin to cut off.
    You are used to common emitter stages, where the emitter is grounded, and the load is in the collector circuit.
     
  12. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    AH! I get that bit now.
    But from collector to base, does the transistor junction drop the (6-3.7=) 2.3 volts?
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Your arithmetic is good.:)
     
  14. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    :)
    The emitter will maintain a voltage 0.6 or 0.7v below the base voltage. So you force base to be 3.7v, it forces emitter to output 3.0v (give or take 0.1v or so).

    The collector voltage is fairly irrelevant, anything between about 3.5v and 30v will work. Obviously the transistor will get hot (like any voltage regulator) if you have too much voltage and/or too much current.

    You BC337 is pretty ideal for a 115mA regulator.
     
  15. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    OK, thanks, I just wasn't sure if it was good for the CB junction, because I know trying to drop too much voltage with a diode is bad.
     
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